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3 A - 3B, g2E - F, H 1/3 F + E, g2A - B - C, DC - D

Mathematics

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Haya, tufanye hesabu za matriki.

Kutokana na picha, tuna matriki A na B kama ifuatavyo:

A=[10651]A = \begin{bmatrix} 10 & 6 \\ 5 & 1 \end{bmatrix} B=[0−6−48]B = \begin{bmatrix} 0 & -6 \\ -4 & 8 \end{bmatrix}

Tutatatua swali la kwanza: 12A−3B\frac{1}{2}A - 3B.

  1. Pata 12A\frac{1}{2}A.

    Zidisha kila kipengele cha matriki A kwa 12\frac{1}{2}.

    12A=12[10651]=[102625212]=[535212]\frac{1}{2}A = \frac{1}{2} \begin{bmatrix} 10 & 6 \\ 5 & 1 \end{bmatrix} = \begin{bmatrix} \frac{10}{2} & \frac{6}{2} \\ \frac{5}{2} & \frac{1}{2} \end{bmatrix} = \begin{bmatrix} 5 & 3 \\ \frac{5}{2} & \frac{1}{2} \end{bmatrix}
  2. Pata $3B$.

    Zidisha kila kipengele cha matriki B kwa 3.

    3B=3[0−6−48]=[3×03×(−6)3×(−4)3×8]=[0−18−1224]3B = 3 \begin{bmatrix} 0 & -6 \\ -4 & 8 \end{bmatrix} = \begin{bmatrix} 3 \times 0 & 3 \times (-6) \\ 3 \times (-4) & 3 \times 8 \end{bmatrix} = \begin{bmatrix} 0 & -18 \\ -12 & 24 \end{bmatrix}
  3. Kokotoa 12A−3B\frac{1}{2}A - 3B.

    Toa vipengele vinavyolingana vya 3B3B kutoka 12A\frac{1}{2}A.

    12A−3B=[535212]−[0−18−1224]\frac{1}{2}A - 3B = \begin{bmatrix} 5 & 3 \\ \frac{5}{2} & \frac{1}{2} \end{bmatrix} - \begin{bmatrix} 0 & -18 \\ -12 & 24 \end{bmatrix} =[5−03−(−18)52−(−12)12−24]= \begin{bmatrix} 5 - 0 & 3 - (-18) \\ \frac{5}{2} - (-12) & \frac{1}{2} - 24 \end{bmatrix} =[53+1852+24212−482]= \begin{bmatrix} 5 & 3 + 18 \\ \frac{5}{2} + \frac{24}{2} & \frac{1}{2} - \frac{48}{2} \end{bmatrix} =[521292−472]= \begin{bmatrix} 5 & 21 \\ \frac{29}{2} & -\frac{47}{2} \end{bmatrix}

    Jibu la kwanza ni:

    [521292−472]\boxed{\begin{bmatrix} 5 & 21 \\ \frac{29}{2} & -\frac{47}{2} \end{bmatrix}}

    Kwa maswali mengine (kama vile 2E−F2E - F, 13F+E\frac{1}{3}F + E, 2A−B−C2A - B - C, na C−DC - D), matriki C, D, E, na F hazionekani kikamilifu kwenye picha. Tafadhali tuma picha kamili ya matriki hizo ili niweze kuzitatua.

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